Where is the Error in sorting code

Question

Hey everybody please help I can't find the error

when I press sort...it sort again and again forever

html:

<div class="unsorted">
    <div class="labelimagediv" rel="5"> 
      Div 5 here 
    </div> 
    <div class="labelimagediv" rel="3"> 
       Div 3 Here 
    </div> 
     <div class="labelimagediv" rel="4.5"> 
       Div 4.5 here 
    </div>
    <div class="labelimagediv" rel="4">
      Div 4 Here
     </div> </div>
<button class="btnSort">Sort</button>

javascript:

$("button.btnSort").click(function () {
    var item1 = $('div.labelimagediv')[0];
    var sortedDivs = $(".unsorted").find(item1).toArray().sort(sorter);
    $.each(sortedDivs, function (index, value) {
        $(".unsorted").append(value);
    });
});

function sorter(a, b) {
    return a.getAttribute('rel') - b.getAttribute('rel');
};

try it jsfiddle.net/NBKPk when I press the sort botton it still sorting and I want only to sort from 1 to 3 one time only — Dave Mas
– Dave Mas, Commented Sep 12, 2013 at 22:12
At first glance, it is NOT sorting, just moving 1 element. The error seems to be in between the keyboard and the chair. — Milche Patern
– Milche Patern, Commented Sep 12, 2013 at 22:16
I'm not like you "programmer" I'm trying to make my things by myself — Dave Mas
– Dave Mas, Commented Sep 12, 2013 at 22:23

Milche Patern · Accepted Answer · 2013-09-12 22:15:06Z

4

That's because you are sorting one element by selecting the first element $('div.labelimagediv')[0];, try this:

$("button.btnSort").click(function () {
    $('div.labelimagediv').sort(function(a, b) {
       return a.getAttribute('rel') > b.getAttribute('rel');
    }).appendTo(".unsorted");
});

JsFiddle fork here

edited Sep 12, 2013 at 22:15

Milche Patern

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answered Sep 12, 2013 at 22:11

Ram

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Comments

adeneo · Accepted Answer · 2013-09-12 22:12:05Z

1

var item1 = $('div.labelimagediv')[0];

^^ that's just one single element ?

It should be :

$("button.btnSort").click(function () {
    var sortedDivs = $('.labelimagediv', '.unsorted').get();
    sortedDivs.sort(function(a,b) {
        return a.getAttribute('rel') - b.getAttribute('rel');
    });
    $.each(sortedDivs, function (index, value) {
        $(".unsorted").append(value);
    });
});

FIDDLE

answered Sep 12, 2013 at 22:12

adeneo

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